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Question

Find the magnitude, in radians and degrees, of the interior angle of a regular (i) pentagon (ii) octagon (iii) heptagon (iv) duodecagon.

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Solution

(i)

Sum of the interior angles of the polygon =n-2πNumber of sides in the pentagon=5 Sum of the interior angles of the pentagon =5-2π=3πEach angle of the pentagon =Sum of the interior angles of the polygonNumber of sides=3π5radEach angle of the pentagon=3π5×180π°=108°

(ii)

Sum of the interior angles of the polygon=n-2πNumber of sides in the octagon=8 Sum of the interior angles of the octagon =8-2π=6πEach angle of the octagon =Sum of the interior angles of the polygonNumber of sides=6π8 =3π4 radEach angle of octagon=3π4×180π°=135°

(iii)

Sum of the interior angles of the polygon=n-2πNumber of sides in the heptagon=7 Sum of the interior angles of the heptagon =7-2π=5πEach angle of the heptagon =Sum of the interior angles of the polygonNumber of sides=5π7 radEach angle of the heptagon=5π7×180π°=9007°=128°34'17''

(iv)

Sum of the interior angles of the polygon =n-2πNumber of sides in the duodecagon=12 Sum of the interior angles of the duodecagon =12-2π=10πEach angle of the duodecagon =Sum of the interior angles of the polygonNumber of sides=10π12 =5π6 radEach angle of duodecagon=5π6×180π°=150°

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